Chapter 10, Problem 40AP

(a)

To determine

The total torques due to the weight of the hand about the axis of rotation when the time reads $3:00$, $5:15$, $6:00$, $8:20$ and $9:45$.

Answer to Problem 40AP

The total torques due to the weight of the hand about the axis of rotation when the time reads $3:00$, $5:15$, $6:00$, $8:20$ and $9:45$ are $794\text{N}\cdot \text{m}$, $2510\text{N}\cdot \text{m}$, $0$, $1160\text{N}\cdot \text{m}$ and $-2940\text{N}\cdot \text{m}$ respectively.

Explanation of Solution

Given information: The mass of hour hand is $60\text{kg}$, mass of minute hand is $100\text{kg}$, length of hour hand is $2.70\text{m}$, length of minute hand is $4.50\text{m}$.

Formula to calculate the net torque produced by the clock’s hand is,

$\begin{array}{c}\tau =m_{\text{h}}g\left(\frac{L_{\text{h}}}{2}\right)\sin {\theta }_{\text{h}}+m_{\text{m}}g\left(\frac{L_{\text{m}}}{2}\right)\sin {\theta }_{\text{m}}\\ =\frac{g}{2}\left(m_{\text{h}}{L}_{\text{h}}\sin {\theta }_{\text{h}}+m_{\text{m}}{L}_{\text{m}}\sin {\theta }_{\text{m}}\right)\end{array}$ (1)

Here,

$\tau$ is the net torque produce by both hands.

$m_{\text{h}}$ is the mass of the hour hand.

$m_{\text{m}}$ is the mass of the minute hand.

$L_{\text{h}}$ is the mass of the hour hand.

$L_{\text{m}}$ is the mass of the minute hand.

${\theta }_{\text{h}}$ is the angle made by the hour hand at time $t$.

${\theta }_{\text{m}}$ is the angle made by the minute hand at time $t$.

Formula to calculate the angular speed of the hour hands is,

${\omega }_{\text{h}}=\frac{\text{One}\text{revolution}}{\text{Time}\text{taken}\text{by}\text{the}\text{hour}\text{hand}\text{to}\text{complete}\text{one}\text{revolution}}$

Substitute $12\text{h}$ for $\text{time}\text{taken}\text{by}\text{the}\text{hour}\text{hand}\text{to}\text{complete}\text{one}\text{revolution}$ in above equation.

$\begin{array}{c}{\omega }_{\text{h}}=\frac{1\text{rev}}{12\text{h}}\times \frac{2\pi \text{rad}}{1\text{rev}}\\ =\frac{\pi }{6}\text{rad}/\text{h}\end{array}$

Thus, the angular speed of the hour hand is $\frac{\pi }{6}\text{rad}/\text{h}$.

Formula to calculate the angular speed of the minute hands is,

${\omega }_{\text{m}}=\frac{\text{One}\text{revolution}}{\text{Time}\text{taken}\text{by}\text{the}\text{minute}\text{hand}\text{to}\text{complete}\text{one}\text{revolution}}$

Substitute $1\text{h}$ for $\text{time}\text{taken}\text{by}\text{the}\text{minute}\text{hand}\text{to}\text{complete}\text{one}\text{revolution}$ in above equation.

$\begin{array}{c}{\omega }_{\text{m}}=\frac{1\text{rev}}{1\text{h}}\times \frac{2\pi \text{rad}}{1\text{rev}}\\ =2\pi \text{rad}/\text{h}\end{array}$

Thus, the angular speed of the hour hand is $2\pi \text{rad}/\text{h}$.

Let take $t=0$ at $12:00$ clock.

Write the expression for the angular position of the hour hands at time $t$ is,

${\theta }_{\text{h}}={\omega }_{\text{h}}t$

Here,

${\omega }_{\text{h}}$ is the angular speed of the hour hand.

Substitute $\frac{\pi }{6}\text{rad}/\text{h}$ for ${\omega }_{\text{h}}$ in above equation.

${\theta }_{\text{h}}=\left(\frac{\pi }{6}\text{rad}/\text{h}\right)t$

Write the expression for the angular position of the minute hands at time $t$ is,

${\theta }_{\text{m}}={\omega }_{\text{m}}t$

Here,

${\omega }_{\text{m}}$ is the angular speed of the minute hand.

Substitute $2\pi \text{rad}/\text{h}$ for ${\omega }_{\text{m}}$ in above equation.

${\theta }_{\text{m}}=\left(2\pi \text{rad}/\text{h}\right)t$

Substitute $\left(\frac{\pi }{6}\text{rad}/\text{h}\right)t$ for ${\theta }_{\text{h}}$ and $\left(2\pi \text{rad}/\text{h}\right)t$ for ${\theta }_{\text{m}}$ in equation (1).

$\tau =\frac{g}{2}\left(m_{\text{h}}{L}_{\text{h}}\sin \left(\frac{\pi t}{6}\right)+m_{\text{m}}{L}_{\text{m}}\sin \left(2\pi t\right)\right)$

Substitute $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $60\text{kg}$ for $m_{\text{h}}$, $100\text{kg}$ for $m_{\text{m}}$, $2.70\text{m}$ for $L_{\text{h}}$ and $4.50\text{m}$ for $L_{\text{m}}$ to fine the equation for $\tau$.

$\begin{array}{c}\tau =\frac{9.8\text{m}/{\text{s}}^{\text{2}}}{2}\left(\left(60\text{kg}\right)\left(2.70\text{m}\right)\sin \left(\frac{\pi t}{6}\right)+\left(100\text{kg}\right)\left(4.50\text{m}\right)\sin \left(2\pi t\right)\right)\\ =4.9\text{m}/{\text{s}}^{\text{2}}\left(\left(162\text{kg}\cdot \text{m}\right)\sin \left(\frac{\pi t}{6}\right)+\left(450\text{kg}\cdot \text{m}\right)\sin \left(2\pi t\right)\right)\\ =\left(794\text{N}\cdot \text{m}\right)\left(\sin \left(\frac{\pi t}{6}\right)+2.78\sin \left(2\pi t\right)\right)\end{array}$ (2)

When clock shows time $3:00$ that gives time $t=3\text{h}$

Substitute $3\text{h}$ for $t$ in equation (2).

$\begin{array}{c}\tau =\left(794\text{N}\cdot \text{m}\right)\left(\sin \left(\frac{\pi \times 3}{6}\right)+2.78\sin \left(2\pi \times 3\right)\right)\\ =\left(794\text{N}\cdot \text{m}\right)\left(\sin \left(\frac{\pi }{2}\right)+2.78\sin \left(6\pi \right)\right)\\ =794\text{N}\cdot \text{m}\end{array}$

Thus, the net torque is $794\text{N}\cdot \text{m}$.

When clock shows time $5:15$ that gives time

$\begin{array}{l}t=\left(5+\frac{15}{60}\right)\text{h}\\ =\text{5}\text{.25}\text{h}\end{array}$

Substitute $\text{5}\text{.25}\text{h}$ for $t$ in equation (2).

$\begin{array}{c}\tau =\left(794\text{N}\cdot \text{m}\right)\left(\sin \left(\frac{\pi \times \text{5}\text{.25}\text{h}}{6}\right)+2.78\sin \left(2\pi \times \text{5}\text{.25}\text{h}\right)\right)\\ =\left(794\text{N}\cdot \text{m}\right)\left(\sin \left(\frac{5.25\pi }{2}\right)+2.78\sin \left(10.5\pi \right)\right)\\ =2510\text{N}\cdot \text{m}\end{array}$

Thus, the net torque is $2510\text{N}\cdot \text{m}$.

When clock shows time $6:00$ that gives time

$t=6\text{h}$

Substitute $6\text{h}$ for $t$ in equation (2).

$\begin{array}{c}\tau =\left(794\text{N}\cdot \text{m}\right)\left(\sin \left(\frac{\pi \times 6}{6}\right)+2.78\sin \left(2\pi \times 6\right)\right)\\ =\left(794\text{N}\cdot \text{m}\right)\left(\sin \left(\pi \right)+2.78\sin \left(12\pi \right)\right)\\ =0\end{array}$

Thus, the net torque is $0$.

When clock shows time $8:20$ that gives time

$\begin{array}{l}t=\left(8+\frac{20}{60}\right)\text{h}\\ =\text{8}\text{.33}\text{h}\end{array}$

Substitute $\text{8}\text{.33}\text{h}$ for $t$ in equation (2).

$\begin{array}{c}\tau =\left(794\text{N}\cdot \text{m}\right)\left(\sin \left(\frac{\pi \times \text{8}\text{.33}\text{h}}{6}\right)+2.78\sin \left(2\pi \times \text{8}\text{.33}\text{h}\right)\right)\\ =\left(794\text{N}\cdot \text{m}\right)\left(\sin \left(\frac{\text{8}\text{.33}\text{h}\pi }{6}\right)+2.78\sin \left(16.6\pi \right)\right)\\ =1160\text{N}\cdot \text{m}\end{array}$

Thus, the net torque is $1160\text{N}\cdot \text{m}$.

When clock shows time $9:45$ that gives time

$\begin{array}{c}t=\left(9+\frac{45}{60}\right)\text{h}\\ =\text{9}\text{.75}\text{h}\end{array}$

Substitute $\text{9}\text{.75}\text{h}$ for $t$ in equation (2).

$\begin{array}{c}\tau =\left(794\text{N}\cdot \text{m}\right)\left(\sin \left(\frac{\pi \times \text{9}\text{.75}\text{h}}{6}\right)+2.78\sin \left(2\pi \times \text{9}\text{.75}\text{h}\right)\right)\\ =\left(794\text{N}\cdot \text{m}\right)\left(\sin \left(\frac{\text{9}\text{.75}\text{h}\pi }{6}\right)+2.78\sin \left(19.5\pi \right)\right)\\ =-2940\text{N}\cdot \text{m}\end{array}$

Thus, the net torque is $-2940\text{N}\cdot \text{m}$.

Conclusion:

Therefore, the total torques due to the weight of the hand about the axis of rotation when the time reads $3:00$, $5:15$, $6:00$, $8:20$ and $9:45$ are $794\text{N}\cdot \text{m}$, $2510\text{N}\cdot \text{m}$, $0$, $1160\text{N}\cdot \text{m}$ and $-2940\text{N}\cdot \text{m}$ respectively.

(b)

To determine

The all the time nearest to second when total torque about the axis of rotation is zero by solving the transcendental equation.

Answer to Problem 40AP

The time corresponding to the zero torque is given as:

Time(hr)	Clock time
0	12:00:00
0.515	12:30:55
0.971	12:58:19
1.54	1:32:31
1.95	1:57:01
2.56	2:33:25
2.94	2:56:29

Explanation of Solution

Given information: The mass of hour hand is $60\text{kg}$, mass of minute hand is $100\text{kg}$, length of hour hand is $2.70\text{m}$, length of minute hand is $4.50\text{m}$.

From equation (2), the expression for the total torque is given by,

$\tau =\left(794\text{N}\cdot \text{m}\right)\left(\sin \left(\frac{\pi t}{6}\right)+2.78\sin \left(2\pi t\right)\right)$

Substitute $0$ for $\tau$ to find all the times.

$\begin{array}{c}\left(794\text{N}\cdot \text{m}\right)\left(\sin \left(\frac{\pi t}{6}\right)+2.78\sin \left(2\pi t\right)\right)=0\\ \sin \left(\frac{\pi t}{6}\right)+2.78\sin \left(2\pi t\right)=0\end{array}$

Since it is a transcendental equation, solving the equation numerically the values of time comes out to be 0, 0.515, 0.971, 1.54, 1.95……so on.

The time corresponding to the time is given as:

Time(hr)	Clock time
0	12:00:00
0.515	12:30:55
0.971	12:58:19
1.54	1:32:31
1.95	1:57:01
2.56	2:33:25
2.94	2:56:29

Conclusion:

Therefore, time corresponding to the zero torque is given as:

Time(hr)	Clock time
0	12:00:00
0.515	12:30:55
0.971	12:58:19
1.54	1:32:31
1.95	1:57:01
2.56	2:33:25
2.94	2:56:29